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Question Number 132928 by metamorfose last updated on 17/Feb/21

Σ_(k=1) ^(+∞) (−1)^k ln(1+(1/k))

$$\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right) \\ $$

Commented by Olaf last updated on 17/Feb/21

sorry sir, I deleted my answer. it was wrong.

$${sorry}\:{sir},\:{I}\:{deleted}\:{my}\:{answer}. \\ $$$${it}\:{was}\:{wrong}. \\ $$

Answered by mnjuly1970 last updated on 17/Feb/21

S=Σ_(k=1) ^n (−1)^k ∫_0 ^( 1) (dx/(x+k)) =∫_0 ^( 1) Σ_(k=1) ^∞ (−1)^k (1/(x+k))dx =_(convergent) ^(conditional) (1/2)∫_0 ^( 1) Σ(−1)^k (1/((x/2)+(k/2)))dx =(1/2)∫_0 ^( 1) {Σ_(k=) ^∞ (1/((x/2)+k)) −Σ_(k=1) ^∞ (1/((x/2)−(1/2)+k))}dx =(1/2)∫_0 ^( 1) {−γ+Σ(1/k)−(1/((x/2)−(1/2)+k))−(−γ+Σ(1/k)−(1/((x/2)+k)))}dx =(1/2)∫_0 ^( 1) {𝛙((x/2)+(1/2))−𝛙((x/2)+1)}dx note: 𝛙(z+1)=−𝛄+Σ_(k=1) ^∞ ((1/k)−(1/(k+z))) ∴ S=(1/2)(2ln(𝚪((x/2)+(1/2)))−2ln(𝚪((x/2)+1))_0 ^1 =−ln𝚪((3/2))−ln(𝚪((1/2)))= =−ln((1/2)(√π) .(√π) )=ln((2/π))...✓✓✓ =∫_0 ^( 1) ((1−x)/((1+x)ln(x)))dx .m.n.july.1970.

$$\:\:{S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dx}}{{x}+{k}} \\ $$$$\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{{x}+{k}}{dx} \\ $$$$\:\underset{{convergent}} {\overset{{conditional}} {=}}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \Sigma\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\frac{{x}}{\mathrm{2}}+\frac{{k}}{\mathrm{2}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\underset{{k}=} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\frac{{x}}{\mathrm{2}}+{k}}\:−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+{k}}\right\}{dx} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{−\gamma+\Sigma\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+{k}}−\left(−\gamma+\Sigma\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{\frac{{x}}{\mathrm{2}}+{k}}\right)\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\boldsymbol{\psi}\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\boldsymbol{\psi}\left(\frac{{x}}{\mathrm{2}}+\mathrm{1}\right)\right\}{dx} \\ $$$${note}:\:\boldsymbol{\psi}\left({z}+\mathrm{1}\right)=−\boldsymbol{\gamma}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+{z}}\right) \\ $$$$\therefore\:{S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{ln}\left(\boldsymbol{\Gamma}\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)−\mathrm{2}{ln}\left(\boldsymbol{\Gamma}\left(\frac{{x}}{\mathrm{2}}+\mathrm{1}\right)\right)_{\mathrm{0}} ^{\mathrm{1}} \right. \\ $$$$=−{ln}\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−{ln}\left(\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)= \\ $$$$=−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}\:.\sqrt{\pi}\:\right)={ln}\left(\frac{\mathrm{2}}{\pi}\right)...\checkmark\checkmark\checkmark \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}}{\left(\mathrm{1}+{x}\right){ln}\left({x}\right)}{dx}\:\:\:\:\:.{m}.{n}.{july}.\mathrm{1970}. \\ $$$$\:\: \\ $$

Commented by metamorfose last updated on 17/Feb/21

thanks , it′s an intresting solution

$${thanks}\:,\:{it}'{s}\:{an}\:{intresting}\:{solution} \\ $$

Commented by mnjuly1970 last updated on 17/Feb/21

grateful sir .....

$$\:{grateful}\:{sir}\:..... \\ $$$$ \\ $$

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